Problem: A bag contains $6$ red jelly beans, $4$ green jelly beans, and $4$ blue jelly beans. If we choose a jelly bean, then another jelly bean without putting the first one back in the bag, what is the probability that the first jelly bean will be green and the second will be red?
The probability of event A happening, then event B, is the probability of event A happening times the probability of event B happening given that event A already happened. In this case, event A is picking a green jelly bean and leaving it out. Event B is picking a red jelly bean. Let's take the events one at at time. What is the probability that the first jelly bean chosen will be green? There are $4$ green jelly beans, and $14$ total, so the probability we will pick a green jelly bean is $\dfrac{4} {14}$. After we take out the first jelly bean, we don't put it back in, so there are only $13$ jelly beans left. Since the first jelly bean was green, there are still $6$ red jelly beans left. So, the probability of picking a red jelly bean after taking out a green jelly bean is $\dfrac{6} {13}$. Therefore, the probability of picking a green jelly bean, then a red jelly bean is $\dfrac{4}{14} \cdot \dfrac{6}{13} = \dfrac{24}{182} = \dfrac{12}{91}$